2015 Cambridge Entrance Examination Q6

Question

Show : $\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{2}{1+\sin(x)}\\$. Then evaluate$\int\frac{dx}{1+\sin(x)}$
Show : $\int_{0}^{\pi}x \space f(\sin(x)) \space dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x)) \space dx$. Then evaluate $\int_{0}^{\pi}\frac{x}{1+\sin(x)}$
Evaluate: $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}dx$

The first trigonometric identity is quite easy to prove. $$\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{1}{\cos^2\left(\frac{1}{2}\left((\frac{\pi}{2}-x\right)\right)}=\frac{1}{1/2\left(1+\cos\left(\frac{\pi}{2}-x\right)\right)}=\frac{2}{1+\sin(x)}$$

And the integral just involves substitution the integrand with the secant term, which alloys us to take advantage of the fact that the derivative of the tangent is secant squared.

The first second integral were tricky at first but allowing $y=\pi-x$ did the trick. However evaluting the integral in the third part...

$$\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}$$

... wasn't as simple as I thought. I'm really not sure where to begin or how to utelise the integral properties proved in the previous parts. Thanks

Use kings rule and add (similar to what you did in second question), and you will be left with an easy integral. — , Jan 18 '22 at 07:30
There could be a number of methods to solve this problem but there is no doubt that the question is phrased so beautifully. Nice $\to +1$ — Darshan P., Jan 18 '22 at 08:18

Darshan P. · Answer 1 · 2022-01-20T08:43:17.947

2

The beauty of this problem is that it used an allied function $\sin(x)$ with phase $\pi$

If you could see the graph of $y = E(x)\times A(x)$ it can help you understand the complexity of the problems

Basically if $E(x)$ is an odd function then this can be solved:

Solution

$\int_0^\pi x^nf(\sin x)dx$ can be solved if $n $ is an odd

Given: $\int_0^\pi x f(\sin x) dx = \frac \pi2\int_0^\pi f(\sin x)dx$

To find $\int_0^\pi \frac {2x^3-3\pi x^2}{(1+\sin x)^2}dx = \int_0^\pi (2x^3-3\pi x^2) f(\sin x)$

Take $$\begin{align*} \int_0^\pi x^3 f(\sin x)dx & = \int_0^\pi(\pi-x)^3f(\sin x)dx\\ & \text { by expansion.......and using the given property }\\ & \implies \int_0^\pi x^3f(\sin x)dx = \int_0^\pi \left(\frac {\pi^3}{2}-\frac{3\pi^3}4 + \frac {3\pi x^2}{2} \right)f(\sin x)dx \end{align*}$$

Coming back to our original problem:

$$I = \int_0^\pi \frac {2\left(-\frac {\pi^3}{2} + \color{blue}{\frac {3\pi x^2}{2}}\right) - \color{blue}{3\pi x^2}}{(1+\sin x)^2}dx$$

Now, I believe you can solve $\int_0^\pi\text {constant }\frac {1}{(1+\sin x)^2}dx$

edited Jan 20 '22 at 08:43

answered Jan 18 '22 at 08:09

Darshan P.

1,181

Being sick for the past few days I may not be able to write the complete solution but if you still find difficulty in understanding I'll try to add the complete solution. Enjoy:-)) – Darshan P. Jan 18 '22 at 08:10
I would suggest you to use as follows $$\int_0^\pi \frac {2x^3}{(1+\sin x)^2}dx - \int_0^\pi\frac {3\pi x^2}{(1+\sin x)^2}dx$$ Now, use $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ for only left part and then use our given identity $\int_0^\pi x f(\sin x)dx = \frac {\pi}2 \int_0^\pi f(\sin x)dx$ rest you can solve. – Darshan P. Jan 18 '22 at 08:16
5

$$\begin{align} I & = \color{blue}{\int_0^\pi x^3f}\ & = \int_0^\pi(\pi - x)^3 f\ & = \int_0^\pi (\pi^3 - x^3 - 3\pi^2x + 3\pi x^2)f\ & = \pi^3\int_0^\pi f -3\pi^2\color{green}{\int_0^\pi x f} + \color{red}{3\pi \int_0^\pi x^2 f} - \color{blue}{\int_0^\pi xf}\ &\implies \color{blue}{2\int_0^\pi x^3 f} - \color{red}{3\pi \int_0^\pi x^2 f} = \pi^3\int_0^\pi f - 3\pi^2\color{green}{\frac \pi{2}\int_0^\pi f} = -\frac 12\pi^3\int_0^\pi f \end{align}$$ – Darshan P. Jan 18 '22 at 08:55
Please let me know if the commented answer is better than the answer itself cz I'm too lazy to write $f(\sin x) = f$ for me – Darshan P. Jan 20 '22 at 08:39

2015 Cambridge Entrance Examination Q6

1 Answers1